$A=\left[\begin{array}{rr}-3 & 3 & -6 & 5 & 8 \\12 & 35 & 13 & 14 & 2 \\-6 &8 & 9 & -18 & 1 \\-5 &-3 &-23 & 6 & 4 \\4 &-5 &15 & 3 & 10\end{array}\right]$ $A_{4,1}=$
Answer: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{4,1}$ $A_{{4},{1}}$ is located on row ${4}$ of $A$ : $\left[\begin{array}{rr}-3 & 3 & -6 & 5 & 8 \\12 & 35 & 13 & 14 & 2 \\-6 &8 & 9 & -18 & 1 \\{-5} & {-3} & {-23} & {6} & {4} \\4 &-5 &15 & 3 & 10\end{array}\right]$ $A_{{4},{1}}$ is also located on column ${1}$ of $A$. $\left[\begin{array}{rr}{-3} & 3 & -6 & 5 & 8 \\{12} & 35 & 13 & 14 & 2 \\{-6} &8 & 9 & -18 & 1 \\{\text{-5}} & {-3} & {-23} & {6} & {4} \\{4} &-5 &15 & 3 & 10\end{array}\right]$ Therefore, $A_{{4},{1}}={-5}$. Summary $A_{4,1}=-5$